Inner product/K/Introduction/Section

In ${ℝ}^n$, we can add vectors and multiply them with a scalar. Moreover, a vector has a certain length, and the relation between two vectors is expressed by the angle between them. Length and angle can be made precise with the concept of an inner product. In order to introduce this, a real vector space or a complex vector space must be given. We want to discuss both cases in parallel, and we use the symbol $𝕂$ to denote $ℝ$ or $ℂ$. For $z\in ℂ$, $\bar{z}$ means the complex-conjugated number; for $z\in ℝ$, this is just the number itself.

Definition

Let $V$ be a $𝕂$-vector space. An inner product on $V$ is a mapping

$V\times V⟶𝕂,(v,w)⟼⟨v,w⟩,$

satisfying the following properties:

1. We have

   $⟨{\lambda }_1{x}_1+{\lambda }_2{x}_2,y⟩={\lambda }_1⟨x_1,y⟩+{\lambda }_2⟨x_2,y⟩$

   for all ${\lambda }_1,{\lambda }_2\in 𝕂$, $x_1,x_2,y\in V$, and

   $⟨x,{\lambda }_1{y}_1+{\lambda }_2{y}_2⟩=\overline{{\lambda }_1}⟨x,y_1⟩+\overline{{\lambda }_2}⟨x,y_2⟩$

   for all ${\lambda }_1,{\lambda }_2\in 𝕂$, $x,y_1,y_2\in V$.

2. We have

   $⟨v,w⟩=\overline{⟨w,v⟩}$

   for all $v,w\in V$.

3. We have $⟨v,v⟩\ge 0$ for all $v\in V$, and $⟨v,v⟩=0$ if and only if $v=0$ holds.

The used properties are called, in the real case, bilinear (which is just another name for multilinear, when we are dealing with the product of two vector spaces), symmetry and positive-definiteness. In the complex case, we call the properties sesquilinear and hermitian. This looks a bit complicated at first sight, but is necessary to ensure that also in the complex case we get positive definiteness and a reasonable concept of distance.

Example

On ${ℝ}^n$, the mapping

${ℝ}^n\times {ℝ}^n⟶ℝ,(v,w)=((v_1,\dots ,v_n),(w_1,\dots ,w_n))⟼\sum _{i=1}^n{v}_i{w}_i,$

defines an inner product, which is called the standard inner product. Simple computations show that this is indeed an inner product.

For example, in ${ℝ}^3$, endowed with the standard inner product, we have

$⟨\left(\begin{array}{c}3\\ -5\\ 2\end{array}\right),\left(\begin{array}{c}-1\\ 4\\ 6\end{array}\right)⟩=3\cdot (-1)-5\cdot 4+2\cdot 6=-11.$

Definition

A real, finite-dimensional vector space that is endowed with an inner product,

is called a Euclidean vector space.

For a vector space $V$, endowed with an inner product, every linear subspace $U\subseteq V$ has again an inner scalar product, by restricting the inner product from $V$. In particular, for a Euclidean vector space, every linear subspace $U\subseteq V$ is again a Euclidean vector space. Therefore, every linear subspace $U\subseteq {ℝ}^n$ carries the restricted standard inner product. Because there is always an isomorphism $U\cong {ℝ}^m$, we can also transfer the standard inner product from ${ℝ}^m$ to $U$. However, the result depends on the isomorphism chosen, and there is, in general, no relationship with the restricted standard inner product.

Definition

On ${ℂ}^n$, we call the inner product given by

$⟨u,v⟩:=\sum _{i=1}^n{u}_i\overline{v_i}$

the

(complex) standard inner product.

For example, we have

$\begin{array}{cc}⟨\left(\begin{array}{c}4-3\mathrm{i}\\ 2+7\mathrm{i}\end{array}\right),\left(\begin{array}{c}-2+5\mathrm{i}\\ 3-6\mathrm{i}\end{array}\right)⟩& =(4-3\mathrm{i})\cdot \overline{-2+5\mathrm{i}}+(2+7\mathrm{i})\cdot \overline{3-6\mathrm{i}}\\ & =(4-3\mathrm{i})\cdot (-2-5\mathrm{i})+(2+7\mathrm{i})\cdot (3+6\mathrm{i})\\ & =-23-14\mathrm{i}-36+33\mathrm{i}\\ & =-59+19\mathrm{i}.\end{array}$

Remark

If we consider a complex vector space $V$, endowed with an inner product $⟨-,-⟩$, as a real vector space, then the real part

$\mathrm{Re}\left(⟨v,w⟩\right)$

is a real inner product, see exercise. Because of

$\begin{array}{cc}⟨v,w⟩& =\mathrm{Re}\left(⟨v,w⟩\right)+\mathrm{i}\mathrm{Im}\left(⟨v,w⟩\right)\\ & =\mathrm{Re}\left(⟨v,w⟩\right)-\mathrm{i}\mathrm{Re}\left(\mathrm{i}⟨v,w⟩\right)\\ & =\mathrm{Re}\left(⟨v,w⟩\right)-\mathrm{i}\mathrm{Re}\left(⟨iv,w⟩\right),\end{array}$

we can reconstruct from the real part the original inner product.

Example

Let $[a,b]$ be a closed real interval with $a<b$, and let

$V=\{f:[a,b]\to ℂ\mid f\text{ continuous}\},$

endowed with pointwise addition and scalar multiplication. Setting

$⟨f,g⟩:={\int }_a^{b}f(t)\overline{g(t)}dt,$

we obtain an inner product on $V$. Additivity follows from

$⟨f_1+f_2,g⟩={\int }_a^b(f_1(t)+f_2(t))\overline{g(t)}dt={\int }_a^b{f}_1(t)\overline{g(t)}dt+{\int }_a^b{f}_2(t)\overline{g(t)}dt=⟨f_1,g⟩+⟨f_2,g⟩.$

It is also positive definite: If $f$ is not the zero function, then let $s\in [a,b]$ denote a point with $f(s)\ne 0$. Therefore, $|f(s)|>0$, and due to the continuity of $f$, there exists a neighborhood $[s-ϵ,s+ϵ]\cap [a,b]$ of length $\ge ϵ$ where

$|f(t)|\ge c>0$

holds for some $c\in {ℝ}_{+}$ (one can take $c=\frac{|f(s)|}{2}$). Hence,

$⟨f,f⟩={\int }_a^b|f(t){|}^{2}dt\ge {\int }_{max(a,s-ϵ)}^{min(b,s+ϵ)}|f(t){|}^{2}dt\ge ϵc^2>0$

is positive.